The only value of x for which {2^{sin x}} + { | vedclass

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Question

(a) Since $A.M.$ $ \ge $ $G.M.$

$\Rightarrow$  $\frac{1}{2}({2^{\sin x}} + {2^{\cos x}}) \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $

$ \Right

Vedclass · Accepted Answer

$\frac{{5\pi }}{4}$

Vedclass · Answer

(a) Since $A.M.$ $ \ge $ $G.M.$

$\Rightarrow$  $\frac{1}{2}({2^{\sin x}} + {2^{\cos x}}) \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $

$ \Rightarrow $ ${2^{\sin x}} + {2^{\cos x}} \ge {2.2^{\frac{{\sin x + \cos x}}{2}}}$

$ \Rightarrow $${2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \frac{{\sin x + \cos x}}{2}}}$

and we know that $\sin x + \cos x \ge - \sqrt 2 $

$	herefore $ ${2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}$, for $x = \frac{{5\pi }}{4}$.

The only value of $x$ for which ${2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}$ holds, is

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